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Question Number 16360 by ajfour last updated on 21/Jun/17

Answered by ajfour last updated on 21/Jun/17

To find x=CD, y=BD, z=AD BE is drawn ⊥ to AD, AF is drawn ⊥ to AD produced. △BDE ∼ △ADF (right ∠ and vertically opposite angles) (y/z)=((BD)/(AD))= ((BE)/(AF)) =((asin θ)/(bsin φ)) ....(i) And y+z = c ....(ii) where c=(√(a^2 +b^2 −2abcos (θ+φ))) so using (i) and (ii), y= (((asin 𝛉)/(asin 𝛉+bsin 𝛗)))c z=(((bsin 𝛗)/(asin 𝛉+bsin 𝛗)))c Now x=CD=CE+DE =CF−DF x = acos θ+asin θ tan δ Also x = bcos φ−bsin φ tan δ ⇒ ((x−acos θ)/(x−bcos φ)) = − ((asin θ)/(bsin φ)) or x(asin θ+bsin φ)=absin θcos φ +abcos θsin φ i.e. x= ((absin (𝛉+𝛗))/(asin 𝛉+bsin 𝛗)) Also cos 𝛅 = ((asin θ)/y) = ((bsin φ)/z) = ((asin 𝛉+bsin 𝛗)/c) .

$$\:{To}\:{find}\:{x}={CD},\:\:{y}={BD},\:{z}={AD} \\ $$$${BE}\:{is}\:{drawn}\:\bot\:{to}\:{AD},\:{AF}\:{is} \\ $$$${drawn}\:\bot\:{to}\:{AD}\:{produced}. \\ $$$$\:\bigtriangleup{BDE}\:\sim\:\bigtriangleup{ADF}\:\:\left({right}\:\angle\:{and}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:{vertically}\:{opposite}\:{angles}\right) \\ $$$$\:\:\frac{{y}}{{z}}=\frac{{BD}}{{AD}}=\:\frac{{BE}}{{AF}}\:=\frac{{a}\mathrm{sin}\:\theta}{{b}\mathrm{sin}\:\phi}\:\:\:\:\:....\left({i}\right) \\ $$$${And}\:\:{y}+{z}\:=\:{c}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({ii}\right) \\ $$$$\:{where}\:\:{c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\left(\theta+\phi\right)}\: \\ $$$$\:{so}\:{using}\:\left({i}\right)\:{and}\:\left({ii}\right), \\ $$$$\:\:\boldsymbol{{y}}=\:\left(\frac{\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\theta}}{\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\theta}+\boldsymbol{{b}}\mathrm{sin}\:\boldsymbol{\phi}}\right)\boldsymbol{{c}} \\ $$$$\:\:\boldsymbol{{z}}=\left(\frac{\boldsymbol{{b}}\mathrm{sin}\:\boldsymbol{\phi}}{\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\theta}+\boldsymbol{{b}}\mathrm{sin}\:\boldsymbol{\phi}}\right)\boldsymbol{{c}} \\ $$$$\:{Now}\:\:\boldsymbol{{x}}={CD}={CE}+{DE} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={CF}−{DF} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}\:=\:{a}\mathrm{cos}\:\theta+{a}\mathrm{sin}\:\theta\:\mathrm{tan}\:\delta\: \\ $$$${Also}\:\:\boldsymbol{{x}}\:=\:{b}\mathrm{cos}\:\phi−{b}\mathrm{sin}\:\phi\:\mathrm{tan}\:\delta \\ $$$$\Rightarrow\:\:\frac{{x}−{a}\mathrm{cos}\:\theta}{{x}−{b}\mathrm{cos}\:\phi}\:=\:−\:\frac{{a}\mathrm{sin}\:\theta}{{b}\mathrm{sin}\:\phi}\: \\ $$$${or}\:{x}\left({a}\mathrm{sin}\:\theta+{b}\mathrm{sin}\:\phi\right)={ab}\mathrm{sin}\:\theta\mathrm{cos}\:\phi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{ab}\mathrm{cos}\:\theta\mathrm{sin}\:\phi \\ $$$${i}.{e}.\:\:\:\boldsymbol{{x}}=\:\frac{\boldsymbol{{ab}}\mathrm{sin}\:\left(\boldsymbol{\theta}+\boldsymbol{\phi}\right)}{\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\theta}+\boldsymbol{{b}}\mathrm{sin}\:\boldsymbol{\phi}}\: \\ $$$$\:\:{Also}\:\:\mathrm{cos}\:\boldsymbol{\delta}\:=\:\frac{{a}\mathrm{sin}\:\theta}{{y}}\:=\:\frac{{b}\mathrm{sin}\:\phi}{{z}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\theta}+\boldsymbol{{b}}\mathrm{sin}\:\boldsymbol{\phi}}{\boldsymbol{{c}}}\:. \\ $$

Commented by mrW1 last updated on 21/Jun/17

good idea!

$$\mathrm{good}\:\mathrm{idea}! \\ $$

Commented by ajfour last updated on 21/Jun/17

i like your simple comment, sir very appropriate for what i did. Thanks Sir.

$${i}\:{like}\:{your}\:{simple}\:{comment},\:{sir} \\ $$$${very}\:{appropriate}\:{for}\:{what}\:{i}\:{did}. \\ $$$${Thanks}\:{Sir}. \\ $$

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